Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$ For $a,b,c>0$. Prove that$:$
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$
My proof:
We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+c)^2} \geqq 0$$
Where
$g(a,b,c) =\frac{1}{16} \left( a+b \right) ^{2} \left( 2\,a+2\,b-c \right) ^{2} \left( 
a+b-2\,c \right) ^{2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{64} \left( a-b \right) ^{2} \cdot \Big[ \left( 2\,c-a-b \right) ^{3} \left( 
119\,a+119\,b+30\,c \right)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b-2\,c \right) ^{2} \left( 343\,{a}^{2}+346\,ab+343\,{b}^{2}
 \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+24\, \left( 2\,c-a-b \right)  \left( a+b \right)  \left( 16\,{a}^{2}+a
b+16\,{b}^{2} \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+36\, \left( 4\,{a}^{2}-5\,ab+4\,{b}^{2} \right)  \left( a+b \right) ^{
2} \Big] \geqq 0$
which is clearly true for $c=\max\{a,b,c\}$
I wish to see another proof without $uvw$! Thanks for a real lot!
You can see also here.
 A: 
here you go                              .                   
A: I found a proof by Titu's Lemma$:$
Let $$\text{LHS} -\text{RHS} \equiv \frac{g(a,b,c)}{4abc(a+b+c)^2}$$
But we have
\begin{align*} g(a,\,b,\,c) &=\frac{1}{2} \sum\limits_{cyc} c^2(a+b-5c)^2 (a-b)^2-9(a-b)^2(b-c)^2(c-a)^2 \\&\geq \frac{1}{2(a^2+b^2+c^2)} \Big[\sum\limits_{cyc} c^2(a+b-5c)(a-b)\Big]^2-9(a-b)^2(b-c)^2(c-a)^2\\&={\frac { \left( a-b \right) ^{2} \left( b-c \right) ^{2} \left( c-a
 \right) ^{2} \left( 7\,{a}^{2}+50\,ab+50\,ac+7\,{b}^{2}+50\,bc+7\,{c}
^{2} \right) }{2\,(a^2+b^2+c^2)}} \geq 0\end{align*}
So we are done.
A: Here $a,b,c$ replaced for $x,y,z$ for convenience.
For $x,y,z>0$ prove that
\begin{align}
\frac{xy}z+\frac{yz}x+\frac{zx}y+
\frac{81xyz}{4(x+y+z)^2}
&\ge
\tfrac74(x+y+z)
\tag{1}\label{1}
\end{align} 
Using Ravi substitution,
\begin{align} 
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
,
\end{align} 
the inequality \eqref{1} transforms
(omitting lengthy, but straightforward details) into equivalent 
inequality
\begin{align} 
85\,r^2+32\,r\,R+64\,R^2
-15\,\rho^2
&\ge 0
\tag{2}\label{2}
\end{align}
in terms of the semiperimeter $\rho$,
inradius $r$ and circumradius $R$ 
of some valid triangle with the side lengths
\begin{align} 
a&=y+z
,\quad
b=z+x
,\quad
c=x+y
.
\end{align}
Dividing \eqref{2} by $R^2$, we get
\begin{align} 
85\,v^2+32\,v+64
-15\,u^2
&\ge 0
\tag{3}\label{3}
,
\end{align}
where $u=\rho/R$, $v=r/R$.
Now all we have to do is to check \eqref{3}
for all valid triangles, that means for all $v\in[0,\tfrac12]$
and $u(v)\in[u_{\min}(v),u_{\max}(v)]$.
Obviously,
\begin{align} 
85\,v^2+32\,v+64
-15\,u^2
&\ge 
85\,v^2+32\,v+64
-15\,u_{\max}(v)^2
\tag{4}\label{4}
\end{align}
where 
\begin{align}
u_{\max}&=\sqrt{27-(5-v)^2+2\sqrt{(1-2\,v)^3}}
\tag{5}\label{5}
,
\end{align}
and we have \eqref{1} equivalent to
\begin{align}
50v^2-59v+17-15\sqrt{(1-2v)^3}
&\ge 0
\tag{6}\label{6}
,\\
(50v^2-59v+17)^2-15^2((1-2v)^3)
&\ge 0
\tag{7}\label{7}
,\\
(25v-8)^2(1-2v)^2
&\ge 0
\tag{8}\label{8}
,
\end{align}
which holds for all values of $v\in[0,\tfrac12]$,
that is for all valid triangles with the side lengths given by
\eqref{2} and hence, 
\eqref{1} is true
for all real $x,y,z$.
A: BW helps!
Let $a\leq b\leq c$, $b=a+u$ and $c=a+u+v.$
Thus, we need to prove that:
$$9(u^2+uv+v^2)a^4+3(14u^3+21u^2v+3uv^2-2v^3)a^2+$$
$$+(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)a^2+$$
$$+u(56u^4+140u^3v+114u^2v^2+31uv^3+v^4)a+$$
$$+4u^2(2u^2+3uv+v^2)^2\geq0,$$
for which it's enough to prove that:
$$4(u^2+uv+v^2)(73u^4+146u^3v+66u^2v^2-7uv^3+v^4)\geq(14u^3+21u^2v+3uv^2-2v^3)^2$$ or
$$292u^6+876u^5v-85u^4v^2+610u^3v^3+391u^2v^4+8uv^5+20v^6\geq0,$$ which is obvious. 
